Extracting the return type from an overloaded function

Question

I want to extract the return type of a function. Problem is, there are other functions with the same name but different signature, and I can not get C++ to select the appropriate one. I know about std::result_of, but from a few tries I have concluded it suffers from the same problem as well. I have heard about a solution involving decltype as well, but I do not know any specifics.

At the moment I am using template metaprogramming to extract the return type from a function pointer type, which works fine for a limited number of parameters (any non-limited solution?), given that extraction of function pointer type works for unambiguous functions.

#include <iostream>

using namespace std;

//  ----

#define resultof(x)     typename ResultOf<typeof(x)>::Type  //  might need a & before x

template <class T>
class ResultOf
{
    public:
        typedef void Type;      //  might need to be T instead of void; see below
};

template <class R>
class ResultOf<R (*) ()>
{
    public:
        typedef R Type;
};

template <class R, class P>
class ResultOf<R (*) (P)>
{
    public:
        typedef R Type;
};

//  ----

class NoDefaultConstructor
{
    public:
        NoDefaultConstructor (int) {}
};


int f ();
int f ()
{
    cout << "f" << endl;
    return 1;
}

double f (int x);
double f (int x)
{
    cout << "f(int)" << endl;
    return x + 2.0;
}

bool f (NoDefaultConstructor);
bool f (NoDefaultConstructor)
{
    cout << "f(const NoDefaultConstructor)" << endl;
    return false;
}

int g ();
int g ()
{
    cout << "g" << endl;
    return 4;
}

int main (int argc, char* argv[])
{
    if(argc||argv){}

//  this works since there is no ambiguity. does not work without &
//  resultof(&g) x0 = 1;
//  cout << x0 << endl;

//  does not work since type of f is unknown due to ambiguity. same thing without &
//  resultof(&f) x1 = 1;
//  cout << x1 << endl;

//  does not work since typeof(f()) is int, not a member function pointer; we COULD use T instead of void in the unspecialized class template to make it work. same thing with &
//  resultof(f()) x2 = 1;
//  cout << x2 << endl;

//  does not work per above, and compiler thinks differently from a human about f(int); no idea how to make it correct
//  resultof(f(int)) x3 = 1;
//  cout << x3 << endl;

//  does not work per case 2
//  resultof(f(int())) x4 = 1;
//  cout << x4 << endl;

//  does not work per case 2, and due to the lack of a default constructor
//  resultof(f(NoDefaultConstructor())) x5 = 1;
//  cout << x5 << endl;

//  this works but it does not solve the problem, we need to extract return type from a particular function, not a function type
//  resultof(int(*)(int)) x6 = 1;
//  cout << x6 << endl;

}

Any idea what syntax feature am I missing and how to fix it, preferably with a solution that works in a simple way, e.g. resultof(f(int))?

Answer 1

I think that this can be done with decltype and declval:

For example: decltype(f(std::declval<T>())).

Answer 2

It's very hard to inspect an overloaded function name without arguments. You can inspect the return types for overloads that differ in arity -- provided that no arity has more than one overload. Even then, turning a hard error (if/when a given arity does have more than one overload) into SFINAE is a pain as it requires writing a trait just for that particular function(!) since overloaded function names can't be passed as any kind of argument. Might as well require user code to use an explicit specialization...

template<typename R>
R
inspect_nullary(R (*)());

template<typename R, typename A0>
R
inspect_unary(R (*)(A0));

int f();
void f(int);

int g();
double g();

typedef decltype(inspect_nullary(f)) nullary_return_type;
typedef decltype(inspect_unary(f)) unary_return_type;

static_assert( std::is_same<nullary_return_type, int>::value, "" );
static_assert( std::is_same<unary_return_type, void>::value, "" );

// hard error: ambiguously overloaded name
// typedef decltype(inspect_nullary(g)) oops;

Given that you're using C++0x, I feel the need to point out that there is (IMO) never a need to inspect a return type beyond typename std::result_of<Functor(Args...)>::type, and that doesn't apply to function names; but perhaps your interest in this is purely academical.